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NIT1202 Operating Systems

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NIT1202 Operating Systems

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Course Code: NIT1202
University: Victoria University

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Country: Australia

Question:
Write a one-page review on an article about the subject of operating systems that appeared in a recent (since Jan 2016) computing magazine or academic journal. Give a summary of the article, including the primary topic, your own summary of the information presented, and the author’s conclusion. Give your personal evaluation of the article, including topics that made the article interesting to you and its relevance to your own experience.
Answer:
From all the recent research about operating system the understanding gained is summarised below.
In general, operating system is software that is used for communicating with the hardware and allows all the other programs to run on the network. It is a group of files that is needed to boost up the computer (Peter, et. al, 2016). The basic functionality of any gadget is gained through operating system. There are many types of operating system; each system is different as they manage the files and folders differently. Operating system is a core element as it communicates with the hardware and works on a specific application program (Hooker,  Parks & Bunda, 2018). Some of the common operating system used is windows in desktop computers, mac OS for apple gadgets, linux and unix is also used in various personal computers,
Opinion- Operating system is an important as without OS computer and software programs would be useless. In my opinion I would state that, operating system is one of the most important software that runs on a computer (Peter, et. al, 2016). It is software that manages the memory, processes and application that would be running on software’s and hardware’s. Without an operating system a computer is useless.
Summary- Operating system allow user to run any application without having a prior understanding of any platform or coding. Operating system can be found as an intermediate between the application program and the hardware of computer.
User make a request to the application, then future application forwards the request to the operating system and then operating system communicates with hardware and forwards the result to application (Shin,et. al, 2014). Thus, it can be said that application and hardware communicate with each other through the medium of operating system.
Operating system itself is built up of many components. One of the most important components of operating system is kernel that manages all the control over memory. User cannot interact with kernel directly. Apart from that, other component is user interface that interacts with the user directly.
Conclusion- Operating system is software that manages all the resources by providing common services for the computer programs. For an application, operating system offers various services like it allow multiple programs to run on a system at the same time. It also allows memory sharing between various applications. The features of parallel processing are possible due to operating system as it divides the time slot for a program so that many applications can run on a single processor.
Some of the information that is given is:

Job List

Job Number
 

Memory Requested
 

Job A

57K

Job B

920K

Job C

50K

Job D

701K

 

Memory Block list

Memory Block

Memory Block Size

Block 1

 900K  

Block 2

910K

Block 3

200K

Block 4

300K

 
Use the best-fit algorithm to indicate which memory blocks are allocated to each of the four arriving jobs.  
In the best fit, the best available memory block is used to fulfil the request. In the smallest memory block is selected so that it can be used in a maximised way. It works on the concept that memory is used optimally. In this case, problem of internal fragmentation is faced if the memory block that is requested by the processes is less that the allocated memory. If the memory block request is higher than the available slot, algorithm is stopped.
The algorithm that is followed in case of best fit is

Initially number of process and blocks are identified.
Then size of each block is found and process is requested
Best memory block is allocated
The track of wasted memory is kept
Stop

Available memory blocks

900k

910k

200k

300k

Thus, when job A requires 57K it is allocated to BLOCK 3, so that less memory is wasted as compared to other block size.

900k

910k

Job a=57k

143k

300k

Later, when job B arrives and request for 920k, it not fitted in any block as requested size is larger than available size
Job C fits in third block but we cannot use block 3, so its ends up in block 3. As request 50k is completed by using Block 3.

900k

910k

Job a=57k

Job c=50k

93k

300k

At last, job D request 710k which can be completed by using Block 1 and block 2 but block 1 will be used so that least memory is wasted.

Job D=710k

190k

910k

Job a=57k

Job c=50k

93k

300k

 
Use the first-fit algorithm to indicate which memory blocks are allocated to each of the four arriving jobs.   
In case of first fit algorithm, the algorithm works under certain steps that are:

Initially number of processes and number of blocks are identified.
Then size of each block and size of process request is identified.
Then processes are allocated

if (size of block>= process size)
// allocate the process
else
// move to next block

Display the processes with the respective blocks in which they are allocated.
stop

Available memory blocks

900k

910k

200k

300k

Job a request for 57k is allocated to block A of size 900k as size of block is greater than process size.

Job A= 57

863K

910k

200k

300k

Then, Job b requests for 920k which cannot be allocated as size of block is less than the requested size of process.
Later, Job C requests for 50k that could be fitted in the available space of block 1 easily

Job A= 57

Job C=50K

756k

910k

200k

300k

At last, Job d request for 710k that is too large and cannot be fitted in available space of block A, so we need it is fitted in block 2.

Job A= 57k

Job C=50K

756k

Job d=710k

200k

200k

300k

Internal fragmentation and external fragmentation
Fragmentation is way of storing file and information in pieces at different location. This can sometimes be an issue as information go out of order because it is stored in a scattered way. There is a difference between both the terms internal fragmentation deals with portioning the memory block in fixed size. The memory is allocated to processes by dividing them in equal memory blocks (Arlt, Rauchfleisch & Schäfer, 2018). This method of fragmentation is inefficient as excess of memory is wasted. The other form of fragmentation is external fragmentation; it uses the unused memory blocks. The unused memory blocks are assigned to the process so that efficiency is improved and memory is used in an optimized way.
The basic difference between internal and external fragmentation is that internal fragmentation leads to wastage of memory blocks as memory blocks are allocated in a fixed segment. On contrast, external fragmentation refers to allocation of memory block that is unused (Parker & Gorobets, 2015). External fragmentation can take place in the small memory holes that are available while in other case it is not possible. In case of, internal fragmentation memory is partitioned in fixed size. In case of external fragmentation memory is divided in form of variable sized block. The issue faced by internal fragmentation can be eliminated by dynamic allocation method (Parker & Gorobets, 2015). On the other hand, issue faced by external fragmentation can be eliminated by using paging and segmentation.

Internal fragmentation

External fragmentation

In case of internal fragmentation that exist a difference between required memory and allocated memory.

In case of external fragmentation the request of a process is fulfilled by using the unused memory chunks in between.

The main memory is divided into fixed sized blocks without checking the actual size requested by the process

Memory allocation is dynamic

Issue faced by internal fragmentation can be eliminated by dynamic allocation method

Issue faced by external fragmentation can be eliminated by using paging and segmentation.
 

If an option is given to select one of the fragmentations, external fragmentation would be chosen as it allocates the memory dynamically and all the free space is used. It also reduces the overall memory usage by using the concept of paging. The overhead required by external fragmentation is quit less.
If a program has 471 bytes and will be loaded into page frames of 100 bytes each, and the instruction to be used is at byte 132, answer the following questions:
How many pages are needed to store the entire job?  
The total number of pages that is required to store the entire job is 5. That is from page 0 to page 4.
As the page 1 will store frames from 0 to 100
Similarly, page 2 will store frames from 101-200
Similarly, page 3 will store frames from 201-300
Similarly, page 4 will store frames from 301-400
And at last page 5 will store from 401-500.
Thus, 471 bytes will be fitted by using 5 pages
Now, Computing the page number and exact displacement for each of the byte addresses where the data is stored.
It is given that; the instruction to be used is at byte 132.
For finding the displacement at page 0, this is done by multiplying the page frame byte* page number (that is 0)+ displacement of page that is zero for this case
Thus, address is 100*0+0=0
So for, Page 0=Displacement is 0 and address is 0
For finding the displacement at page 1, this is done by multiplying the page frame byte* page number (that is 1)+ displacement of page that is zero for this case as it is not known
Thus, address is 100*1+0=100
So, for Page 1=Displacement is 0 and address is 100
It is given to find instruction to be used is at byte 132.
Thus it will search for the result on the first page.
Therefore,
Address is 132*1+0=132
For, Page 1=Displacement is 32 and address is 132
Page 1 is where instruction is located
For finding the displacement at page 2, this is done by multiplying the page frame byte* page number (that is 2)+ displacement of page that is zero for this case as it is not known
Thus, address is 100*2+0=200
So, for Page 2=Displacement is 0 and address is 200
For finding the displacement at page 3, this is done by multiplying the page frame byte* page number (that is 3)+ displacement of page that is zero for this case as it is not known
thus address is, 100*3+0=300
So, for Page 3, Displacement is 0 and address is 300
For finding the displacement at page 4, this is done by multiplying the page frame byte* page number (that is 0)+ address of page that is 71 for this case
Thus, address is100*4+71=471
So for, Page 4=Displacement 71 and address 400
Given that main memory is composed of only three page frames for public use and that a seven-page program (with pages a, b, c, d, e, f, g) that requests pages in the following order:
a, c, a, b, a, d, a, c, b, d, e, f

Using the FIFO page removal algorithm, indicate the movement of the pages into and out of the available page frames (called a page trace analysis) indicating each page fault with an asterisk (*). Then compute the failure and success ratios.  

Page request

a

c

a

b

a

d

a

c

b

d

e

f

Page fault

*

*

 

*

 

*

 

*

*

 

*

*

Page 1

a

a

a

a

a

a

a

a

b

b

b

f

Page 2

 

c

c

c

c

d

d

d

d

d

d

d

Page 3

 

 

 

b

b

b

b

c

c

c

e

e

Total Faults= 8
Total success= 4
Failure ratio= 8/12
And success ratio= 4/12
Failure percentage =66.66%
Success percentage= 33.33%

Increase the size of memory so it contains four page frames for public use. Using the same page requests as above and FIFO, do another page trace analysis and compute the failure and success ratios.  

Page request

a

c

a

b

a

d

a

c

b

d

e

f

Page fault

*

*

 

*

 

*

 

 

 

 

*

*

Page 1

a

a

a

a

a

a

a

a

a

a

e

e

Page 2

 

c

c

c

c

c

c

c

c

c

c

f

Page 3

 

 

 

b

b

b

b

b

b

b

b

b

Page 4

 

 

 

 

 

d

d

d

d

d

d

d

Total Faults=6
Total success= 6
Failure ratio= 6/12
 And success ratio= 6/12
Failure percentage =50%
Success percentage= 50%/

What general statement can you make from this example? Explain your answer.  

Thus, these questions are solved using first in first out technique. That is the page that is arrived first will be replaced on the new arrival. In the above example the first page is always replaced if new page arrives
Five jobs arrive nearly simultaneously for processing and their estimated CPU cycles are, respectively: Job A = 12, Job B = 2, Job C = 15, Job D = 7, and Job E = 3 ms. (Arabnejad & Barbosa, 2014)

Using FCFS, and assuming the difference in arrival time is negligible, in what order would they be processed? What is the total time required to process all five jobs? What is the average turnaround time for all five jobs?   

Given arrival time of each job = 0ms
So, total time that is taken to complete all eth jobs is, 12+2+15+7+3
Total time needed to complete all the five jobs =39 ms
Turnaround time= Finish time-arrival time
Turnaround time for A is, 12-0=12

                                  A=12

0 12
Turnaround time for B is, 14-0=14

A

B=2

0 12 14
Turnaround time for C is, 29-0=29

A

B

C=15

0 12 14 29
Turnaround time for D is, 36-0=36

A

B

C

D=7

0 12 14 29 36 
Turnaround time for E is, 39-0=39

A

B

C

D

E=3

0    12 14      29               36  39
Average Turnaround time for all the jobs is (12+14+29+36+39)/5= 26 ms
Thus, using FCFS order will be A,B,C,D,E (Qin & Jin, 2015)

Using SJN, and assuming the difference in arrival time is negligible, in what order would they be processed? What is the total time required to process all five jobs? What is the average turnaround time for all five jobs?  

Given arrival time of each job = 0ms
The Turnaround time is difference between finish time and arrival time.
In SJN, shortest job next that is the job is shortest CPU time is processed
Turnaround time for B is, 2-0=2

B=2

0           2
Turnaround time for E is, 5-0=5

B

E=3

0           2 5
Turnaround time for D is, 12-0=12

B

E

D=7

0           2 5 12
Turnaround time for A is, 24-0=24

B

E

D

A=12

0  2 5                    12    24
Turnaround time for C is, 39-0=39

B

E

D

A

C=15

0                   2 5         12  24                         39
Average Turnaround time for all the jobs is (2+5+12+24+39)/5= 16.4ms.
Using SJN, the order will be B,E,D,A,C (Vasile, Pop,  Tutueanu, Cristea & Ko?odziej, 2015)
Given the following information :

Job

Arrival Time

CPU
Cycle

A

0

15

B

2

2

C

3

14

D

6

10

E

9

1

 

Job

Arrival Time

CPU
Cycle

FCFS

SJN

SRT

 
RR

A

0

15

Start:0
Finish:15

Start: 0
Finish:15

Start:0
Finish:28

Start:0
Finish:38

B

2

2

Start:15
Finish:17

Start:16
Finish:18

Start:2
Finish:4

Start:5
Finish:7

C

3

14

Start:17
Finish:31

Start:28
Finish: 42

Start:28
Finish:42

Start:7
Finish:42

D

6

10

Start: 31
Finish:41

Start: 18
Finish:28

Start:6
Finish:17

Start:12
Finish:33

E

9

1

Start: 41
Finish:42

Start:15
Finish:16

Start:9
Finish:10

Start: 17
Finish:18

 

                A=15

0 15

                A

B=2

0 15 17

                A

B

C=14

0 15 17 31

                A

B

C

D=10

0 15 17 31 41

                A

B

C

D

E=1

0 15 17 31 41    51
SJN

                A=15

0 15

                A

E=1

0 15 16

                A

E

B=2

0 15 16 18

                A

E

B

D=10

0 15 16 18 28

                A

E

B

D

C=14

0 15 16       18           28    42
SRT

                A

0 2

A

B

0 2 4

A

B

A

0 2 4 6

A

B

A

D

0 2 4 6 9

A

B

A

D

E

0     2 4      6 9 10

A

B

A

D

E

D

A

C

0     2 4                    6     9 10              17 28    42
RR
Quantum time is 5

A

0 5

A

B

0 5    7          

A

B

C

0 5    7          12

A

B

C

D

0 5    7          12 17

A

B

C

D

E

0 5    7          12 17    18

A

B

C

D

E

A

0 5    7          12 17    18       23

A

B

C

D

E

A

C

0 5    7          12 17    18       23       28    

A

B

C

D

E

A

C

D

0 5    7          12 17    18       23 28 33    

A

B

C

D

E

A

C

D

A

0 5    7          12 17    18       23 28 33    38    

A

B

C

D

E

A

C

D

A

C

0 5    7          12 17    18       23 28 33    38   42
References
Arabnejad, H., & Barbosa, J. G. (2014). List scheduling algorithm for heterogeneous systems by an optimistic cost table. IEEE Transactions on Parallel and Distributed Systems, 25(3), 682-694.
Arlt, D., Rauchfleisch, A., & Schäfer, M. S. (2018). Between Fragmentation and Dialogue. Twitter Communities and Political Debate About the Swiss “Nuclear Withdrawal Initiative”. Environmental Communication, 1-18.
Hooker, R. E., Parks, T., & Bunda, J. D. (2018). U.S. Patent No. 9,891,918. Washington, DC: U.S. Patent and Trademark Office.
Parker, L. M., & Gorobets, S. A. (2015). U.S. Patent No. 8,990,477. Washington, DC: U.S. Patent and Trademark Office.
Peter, S., Li, J., Zhang, I., Ports, D. R., Woos, D., Krishnamurthy, A., … & Roscoe, T. (2016). Arrakis: The operating system is the control plane. ACM Transactions on Computer Systems (TOCS), 33(4), 11.
Qin, L. I. U., & Jin, C. H. E. N. (2015). Job-Shop Scheduling with Parallel Equipments Based on Five Dimensions Scheduling Algorithm. Journal of Jiangnan University (Natural Science Edition), 1, 010.
Shin, S., Song, Y., Lee, T., Lee, S., Chung, J., Porras, P., … & Kang, B. B. (2014, November). Rosemary: A robust, secure, and high-performance network operating system. In Proceedings of the 2014 ACM SIGSAC conference on computer and communications security (pp. 78-89). ACM.
Vasile, M. A., Pop, F., Tutueanu, R. I., Cristea, V., & Ko?odziej, J. (2015). Resource-aware hybrid scheduling algorithm in heterogeneous distributed computing. Future Generation Computer Systems, 51, 61-71.

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